# Find the count of subsequences where each element is divisible by K

• Last Updated : 10 Mar, 2022

Given an array arr[] and an integer K, the task is to find the total number of subsequences from the array where each element is divisible by K.
Examples:

Input: arr[] = {1, 2, 3, 6}, K = 3
Output:
{3}, {6} and {3, 6} are the only valid subsequences.
Input: arr[] = {5, 10, 15, 20, 25}, K = 5
Output: 31

Approach: Since each of the elements must be divisible by K, total subsequences are equal to 2cnt where cnt is the number of elements in the array that are divisible by K. Note that 1 will be subtracted from the result in order to exclude the empty subsequence. So, the final result will be 2cnt – 1.
Below is the implementation of the above approach:

## C++

 // C++ implementation of the approach#include using namespace std; // Function to return the count// of all valid subsequencesint countSubSeq(int arr[], int n, int k){     // To store the count of elements    // which are divisible by k    int count = 0;     for (int i = 0; i < n; i++) {         // If current element is divisible by        // k then increment the count        if (arr[i] % k == 0) {            count++;        }    }     // Total (2^n - 1) non-empty subsequences    // are possible with n element    return (pow(2, count) - 1);} // Driver codeint main(){    int arr[] = { 1, 2, 3, 6 };    int n = sizeof(arr) / sizeof(arr[0]);    int k = 3;     cout << countSubSeq(arr, n, k);     return 0;}

## Java

 // Java implementation of the approachimport java.util.*;class GFG{ // Function to return the count// of all valid subsequencesstatic int countSubSeq(int arr[], int n, int k){     // To store the count of elements    // which are divisible by k    int count = 0;     for (int i = 0; i < n; i++)    {         // If current element is divisible by        // k then increment the count        if (arr[i] % k == 0)        {            count++;        }    }     // Total (2^n - 1) non-empty subsequences    // are possible with n element    return (int) (Math.pow(2, count) - 1);} // Driver codepublic static void main(String[] args){    int arr[] = { 1, 2, 3, 6 };    int n = arr.length;    int k = 3;     System.out.println(countSubSeq(arr, n, k));}} // This code is contributed by Rajput-Ji

## Python3

 # Python3 implementation of the approach # Function to return the count# of all valid subsequencesdef countSubSeq(arr, n, k) :     # To store the count of elements    # which are divisible by k    count = 0;     for i in range(n) :         # If current element is divisible by        # k then increment the count        if (arr[i] % k == 0) :            count += 1;     # Total (2^n - 1) non-empty subsequences    # are possible with n element    return (2 ** count - 1); # Driver codeif __name__ == "__main__" :     arr = [ 1, 2, 3, 6 ];    n = len(arr);    k = 3;     print(countSubSeq(arr, n, k)); # This code is contributed by AnkitRai01

## C#

 // C# implementation of the approachusing System;     class GFG{ // Function to return the count// of all valid subsequencesstatic int countSubSeq(int []arr, int n, int k){     // To store the count of elements    // which are divisible by k    int count = 0;     for (int i = 0; i < n; i++)    {         // If current element is divisible by        // k then increment the count        if (arr[i] % k == 0)        {            count++;        }    }     // Total (2^n - 1) non-empty subsequences    // are possible with n element    return (int) (Math.Pow(2, count) - 1);} // Driver codepublic static void Main(String[] args){    int []arr = { 1, 2, 3, 6 };    int n = arr.Length;    int k = 3;     Console.WriteLine(countSubSeq(arr, n, k));}} // This code is contributed by 29AjayKumar

## Javascript



Output:

3

Time Complexity: O(n)
Auxiliary Space: O(1)

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